In $K(\alpha)/K$, example of different $K$-automorphisms with the same image of $\alpha$

Question

Let $F/K$ be a field extension, and let $\alpha \in F$. If $K(\alpha)/K$ is finite, then a $K$-isomorphism $\sigma \colon K(\alpha) \to K(\alpha)$ is uniquely determined by the image of $\alpha$ (as from that we can deduce the image of an arbitrary element by writing the latter as a polynomial expression in $\alpha$ of finite degree and using the homomorphism properties of $\sigma$).

Is there an example, when $K(\alpha)/K$ is infinite, of $K$-isomorphisms $\sigma, \sigma' \colon K(\alpha) \to K(\alpha)$ with $\sigma(\alpha) = \sigma'(\alpha)$ but $\sigma \ne \sigma'$?

Answer 1

The answer to the original question is no, there are no such examples.

Following Thorgott's hint, we can prove this more general statement:

Let $F/K$ be a field extension, and let $\alpha_1, \ldots, \alpha_n \in F$. Define $K' := K(\alpha_1, \ldots, \alpha_n)$. Then any $K$-automorphism $\sigma \in \mathrm{Gal}(K'/K)$ is uniquely determined by the images of $\alpha_1, \ldots \alpha_n$. That is, if $\sigma' \in \mathrm{Gal}(K'/K)$ and $\sigma'(\alpha_i) = \sigma(\alpha_i)$ for all $i$, then $\sigma' = \sigma$.

Indeed, define the set: $$ A := \{x \in K' \mid \sigma(x) = \sigma'(x)\} \subseteq K' $$ Then:

• $A$ is a subfield of $F$. Indeed:

  • Since by definition of $\mathrm{Gal}(K'/K)$, both $\sigma$ and $\sigma'$ fix $K$ pointwise, we have $1 \in K$ and therefore $1 \in A$. Moreover, since $K$ is a field, $1 \ne 0$, so this means that $A \supsetneq \{0\}$.
  • Let $x, y \in A$. Consider $x - y \in K'$. Then, since $\sigma$ and $\sigma'$ are ring homomorphisms: $$ \sigma(x - y) = \sigma(x) - \sigma(y) = \sigma'(x) - \sigma'(y) = \sigma'(x - y) $$ So $x - y \in A$.
  • Let $x, y \in A$ with $y \ne 0$. Consider $x/y \in K'$. Then, since $\sigma$ and $\sigma'$ are ring homomorphisms: $$ \sigma(x/y) = \sigma(x)/\sigma(y) = \sigma'(x)/\sigma'(y) = \sigma'(x/y) $$ So $x/y \in A$.

  By the three-step subfield test we conclude that $A$ is a subfield of $F$ (so, in particular, $A$ is a field).

• $A$ contains $K$ and $\alpha_1, \ldots, \alpha_n$. Indeed, $K \subseteq A$ is due to the fact that both $\sigma$ and $\sigma'$ fix $K$ pointwise. Also, $\alpha_i \in A$ is true by hypothesis ($\sigma(\alpha_i) = \sigma'(\alpha_i)$) for all $i$.

So, by definition, this implies that $K' = K(\alpha_1, \ldots, \alpha_n) \subseteq A$. The inclusion $K' \supseteq A$ is direct from construction. Therefore $A = K'$. This means that $\sigma$ and $\sigma'$ agree on their entire domain, so $\sigma = \sigma'$.